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Homework Help: Work done by a variable force question

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data

    246kgu9.jpg

    That's a graph of Force in newtons (y axis) vs displacement in meters (x axis) for an object of mass 3.0kg that is moving along the x-axis and initially starts at rest. I am being asked to find the total work done on the object as it moves from x = 0 to x = 7.


    2. Relevant equations

    Net Work = ∫F dx = ΔK
    K = 1/2mv^2

    3. The attempt at a solution

    So, I integrated the function and ended up with a total work of -.5J. What I want to know is if this violates the work-energy theorem or not. If the net work is equal to change in kinetic energy, and the object starts at rest, this would mean that -.5 = 1/2mv^2. Is there something really obvious that I'm missing here?
     
  2. jcsd
  3. Nov 18, 2011 #2
    work done (energy) = area under F against x graph (this is another way of seeing ΔK = ∫F.dx
    There are 2 clear areas to calculate..... when F is +ve and when F is -ve
     
  4. Nov 18, 2011 #3
    By calculating the areas I got total work = -0.5J
     
  5. Nov 18, 2011 #4
    Can you explain why you think the work-energy theorem is violated?

    [itex]0.5=\frac{1}{2}mv^{2}_{f}-\frac{1}{2}mv^{2}_{i}[/itex]

    If the object started at rest, the last term is 0 and you get

    [itex]0.5=\frac{1}{2}mv^{2}_{f}[/itex]

    which just says that the object has some final velocity.
     
  6. Nov 19, 2011 #5
    Right, but if the total work done is -.5, wouldn't that mean that
    [itex]-0.5=\frac{1}{2}mv^{2}_{f}[/itex]
    I'm not sure why you ignored the sign on the work done.
     
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